A proton in a cyclotron changes its velocity from 30 km/s north to 40 km/s east in 20s. What is the magnitude of average acceleration during this time?

Question

A proton in a cyclotron changes its velocity from 30 km/s north to 40 km/s east in 20s. What is the magnitude of average acceleration during this time? (A)  2.5 km/s2  (B)  12.5 km/s2  (C)  22.5 km/s2  (D)  32.5 km/s2

Tardigrade Admin · Accepted Answer

Change in velocity =√(40)2+(30)2 =50 km / s <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ac18dbe27e418e9c6a5df4b45394229e-.png" /> Average acceleration =(50/20) =2.5 km / s 2

Q. A proton in a cyclotron changes its velocity from $30 km / s$ north to $40 km / s$ east in $20 s$ . What is the magnitude of average acceleration during this time?

$2.5 km / s^{2}$

$12.5 km / s^{2}$

$22.5 km / s^{2}$

$32.5 km / s^{2}$

Change in velocity $= (40)^{2} + (30)^{2}$
$= 50 km / s$

Average acceleration $= \frac{50}{20}$
$= 2.5 km / s^{2}$

Q. A proton in a cyclotron changes its velocity from 30km/s north to 40km/s east in 20s. What is the magnitude of average acceleration during this time?

2.5km/s2

12.5km/s2

22.5km/s2

32.5km/s2